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Little Help with a Calc Question

Discussion in 'Pre-Medical - MD' started by americanangel, Apr 22, 2004.

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  1. americanangel

    americanangel Senior Member
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    Okay I'm stuck on this for my calc II class...
    integral of (x^3)(e^-x^2) and integral of (x^3)(2+x)^(5/2)
    I know both are integration by parts but nothing is working!!! Somebody please help!!!
     
    #1 americanangel, Apr 22, 2004
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  2. Athomeonarock

    Athomeonarock Senior Member
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    #2 Athomeonarock, Apr 22, 2004
  3. americanangel

    americanangel Senior Member
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    Great site but i need to show work?? Any ideas how to get the answer?
     
    #3 americanangel, Apr 22, 2004
  4. Cerberus

    Cerberus Heroic Necromancer
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    damn, its been too long since calc two. Looks like integration by parts and substitution though.
     
    #4 Cerberus, Apr 22, 2004
  5. americanangel

    americanangel Senior Member
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    yeah that is what i was thinking but i keep getting these freaky integrals that dont match the real answer!!!

    thanks for the reply

    any ideas welcome!!!
     
    #5 americanangel, Apr 22, 2004
  6. Cerberus

    Cerberus Heroic Necromancer
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    Ok, i'll do the first one.

    int(x^3)(e^-x^2)dx

    let u=x^2

    then

    int[u*x*e^(-u)dx]
    and dx = du*1/(2x)

    so

    int[u*x*e^(-u)(1/(2x)du) = 1/2*int[u*e^(-u)du]

    by integration by parts:

    1/2*int[u*e^(-u)du] = 1/2[-u*e^(-u) - int[1*e^(-u)du] = -1/2*u*e^(-u) + 1/2e^(-u) = -1/2*x^2*e^(-x^2) + 1/2e^(-x^2) = e^(-x^2)*(1-x^2)/2
     
    #6 Cerberus, Apr 22, 2004
  7. americanangel

    americanangel Senior Member
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    oh cool...that found my mistake...i did u=x instead of x^2
    why i did that I dont know but at least I straightened that out!!!

    thanks so much for that!!!
    i really appreciate it!!!
     
    #7 americanangel, Apr 22, 2004
  8. kels

    kels Member
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    For the second one, just substitute u=x+2. So you'll have (u-2)^3*u^(5/2). Expand the first term (u-2)^3 and you'll get something slight messy, but then you can multiply tern by term with u^(5/2), so you'll have a long polynomial with terms like u^(11/2). Then you can integrate those individually, like 12/13*u^(13/12).
     
    #8 kels, Apr 22, 2004
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